= Solution
Differentiate the <Noether charge> and use $\dot\phi_a=\pi_a$ and the <Klein-Gordon equation> for each species. The momentum products cancel because the two species commute:
$$
\begin{aligned}
\dot Q&=\int d^3x\,[\pi_2\pi_1+\phi_2(\nabla^2-m_1^2)\phi_1
-\pi_1\pi_2-\phi_1(\nabla^2-m_2^2)\phi_2]\\
&=\int d^3x\,[\phi_2\nabla^2\phi_1-\phi_1\nabla^2\phi_2]
+(m_2^2-m_1^2)\int d^3x\,\phi_1\phi_2.
\end{aligned}
$$
The first integral is the boundary flux of $\phi_2\nabla\phi_1-\phi_1\nabla\phi_2$, so it vanishes under the same boundary condition used for charge conservation. Hence
$$
\boxed{\frac{dQ}{dt}=(m_2^2-m_1^2)\int d^3x\,\phi_1\phi_2.}
$$
Conservation must hold for arbitrary configurations, rather than only a state in which this particular integral happens to vanish. Thus the <internal rotation symmetry of two real scalar fields> requires
$$
\boxed{m_1^2=m_2^2,\qquad\text{and therefore }m_1=m_2\text{ for nonnegative masses}.}
$$
Equivalently, the local divergence is $\partial_\mu j^\mu=(m_2^2-m_1^2)\phi_1\phi_2$, exhibiting exactly how unequal masses break the symmetry.
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